A counterexample to the Perles conjecture

Electronic Geometry Model No. 2002.04.001

Author

Nikolaus Witte

Description

M. Perles [2] conjectured that the facet subgraphs
of a simple convex d-dimensional polytope are exactly the (d-1)-regular, connected,
induced, non-separating subgraphs. The conjecture does hold for dimension
three and smaller. However, Christian Haase and Günter
M. Ziegler [1] proved that the conjecture does not hold in dimension
four or greater. Based on their ideas we constructed a counterexample
in dimension 4, which, however is smaller than the original construction. Counterexamples in higher
dimensions may be obtained by wedging [1].

The counterexample presented here
(perles.poly) is a simple 4-polytope with 2592 vertices and 529
facets. The 3-regular, connected, induced, non-separating subgraph
which does not correspond to a facet has 1076 nodes. The construction
of Haase and Ziegler [1] (and the one presented here) actually produces a
counterexample to the dual version of the weak Perles conjecture in
dimension 4. Therefore the applet (figure 1) illustrates the
dual case: it depicts a modification of Bing's House, a 2-complex
without a free edge and without 2-dimensional homology. The core idea
of Haase and Ziegler is to embed Bing's House into the boundary
complex of the simplicial counterexample.

Definition: A simplicial d-polytope satisfies
the weak Perles conjecture if the vertex stars are the only pure
(d-1)-dimensional, dually (d-1)-connected, non-separating subcomplexes
of its boundary for which every maximal simplex has one free
facet.

Haase and Ziegler [1] observed that for any (simplicial)
4-dimensional counterexample Q to the dual version of the weak
Perles conjecture, the boundary of Q contains a 2-complex
without a free edge, and without 2-dimensional homology. Examples of
such complexes are known and in the following a modification of
Bing's House is used to construct the counterexample.

This modification of Bing's House consists of two
rooms, the UpperRoom and the LowerRoom,
where each of them is connected to the outside via a `chimney' through
the other room (figure 1). The modified Bing's House can be embedded in a pile of 2*3*4 cubes,
triangulated as follows: Each cube is split into 6 tetrahedra, and the
tetrahedra are all grouped around one of the diagonal axes of the
cube. Add a cone over the boundary with apex w to get a
simplicial 3-sphere C. The
complex C is the boundary complex of a simplicial 4-polytope Q
and C contains a triangulation of the modified Bing's House as a
subcomplex. In the following
C will refer to the whole complex and B will refer to the subcomplex representing
Bing's House. Now the construction proceeds in a
series of stellar subdivisions on edges of C. Essentially these
subdivisions ensure that B becomes an induced subcomplex and
that C is `finely' triangulated: The subcomplex which in the
end will disprove the Perles conjecture will be composed of partial
vertex stars (of vertices in B) covering B. (Each
triangle of B is contained in exactly two of the chosen partial
vertex stars whose
intersection lies entirely in B.) Each tetrahedron of this
subcomplex must have exactly one free triangle, so the triangulation
must be fine enough to ensure that any two of the chosen partial
vertex stars do not intersect outside B.

In order to find the subcomplex in question, Haase and Ziegler [1]
give a coloring of the vertices of B such that the color of a
vertex determines which part of its vertex star (the part lying in
the UpperRoom, the part lying in the LowerRoom or the
part which lies in none of the rooms) is added to the subcomplex.

Finally we have to construct Q, a 4-polytope with the same
combinatorics as C. This process of `lifting' C into
R4 is done as follows:

For a given d-polytope P consider the following
operation: transform P projectively by
moving a hyperplane to `infinity' which separates one vertex w of P from
the others. So, if the halfspace H defined by H := {x in
Rd : ax + ad+1 > 0} contains all vertices
of P except w and is bounded by the hyperplane K :=
{x in Rd : ax + ad+1 = 0}, we define the
projective transformation p as p(x) := x / (ax +
ad+1) , x in Rd\K. This is not an admissible projective transformation for
P, in particular it is not even defined on all of P. The image p(P\K) is a union of two unbounded
polyhedra, the edges
incident to w become rays and the lines they determine all
intersect in p(w). Now we remove the last coordinates from the
vertices in p(P\K) but keep the combinatorial information of
P; we end up with a (d-1)-sphere projected to
Rd-1 with the same combinatorics as P (see
figure 2).

The aim is to reverse this process: Starting with C we try
to find fourth coordinates for its vertices such that we get Q
after a projective transformation with a hyperplane separating the apex
w from the other vertices.

In the construction of C we start
with the triangulated pile of 2*3*4 cubes described above. The
vertices of the cubical complex lie in
Z3. After adding the cone over the boundary with apex
w and completing all stellar
subdivisions the solution of a linear program yields fourth coordinates
for the vertices of C. In this instance, the linear program was
solved with CPLEX and the solution was checked for feasibility
afterwards. The resulting embedding of C in
R4 has the property that after removing all
the facets containing w the remaining complex is embedded on a convex surface U,
all the facets containing w are embedded on a cone (with apex
w) and the cone lies below U. One may imagine this
situation as a bowl U sitting in a cone. In figure 2
the bowl U corresponds to the black lines, the cone to the (dashed and
solid) blue ones. Finally the lifted complex has to be projectively
transformed by moving a hyperplane separating the apex w from
the other vertices to infinity to obtain Q.

At the end of their paper Haase and Ziegler [1] ask whether the
facet subgraphs of a simple 4-polytope are exactly the 3-regular,
3-connected, induced, non-separating and planar
subgraphs. For the example presented here, Petra Mutzel (Technical
University Vienna)
kindly examined the subgraph in question and proved its
non-planarity. This is not surprising, since the subgraph (with
its embedding) is a triangulation of Bing's House. Therefore this
example is not a counterexample to the proposed question.

In the following the content of perles.poly is described
briefly. The two sections which determine the counterexample are
VERTICES and the non standard section SUBGRAPH_NODES containing
the vertices which induce the 3-regular, 3-connected and
non-separating subgraph which does not correspond to a facet. Any
further section can be computed from this information, for example by
polymake, though computation might take some time. Therefore
some additional sections are included in the file. In the following a
complete list of the sections in perles.poly: